Question:

Let $L = \lim_{x \to 0} \left[\frac{3}{2} + \frac{x^2}{\sin x \tan x}\right]$. Step 1: Evaluate the limit of the expression inside the greatest integer function. Consider the term $\frac{x^2}{\sin x \tan x}$. We can rewrite $\tan x$ as $\frac{\sin x}{\cos x}$: $$ \frac{x^2}{\sin x \tan x} = \frac{x^2}{\sin x \left(\frac{\sin x}{\cos x}\right)} = \frac{x^2 \cos x}{\sin^2 x} $$ This can be rearranged as: $$ \frac{x^2}{\sin^2 x} \cdot \cos x = \left(\frac{x}{\sin x}\right)^2 \cdot \cos x $$ Now, we evaluate the limit as $x \to 0$: $$ \lim_{x \to 0} \left[\left(\frac{x}{\sin x}\right)^2 \cdot \cos x\right] $$ We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$, so $\lim_{x \to 0} \frac{x}{\sin x} = 1$. Also, $\lim_{x \to 0} \cos x = \cos 0 = 1$. Therefore, $$ \lim_{x \to 0} \frac{x^2}{\sin x \tan x} = (1)^2 \cdot 1 = 1 $$ Step 2: Substitute this limit back into the original expression. The expression inside the greatest integer function approaches: $$ \frac{3}{2} + 1 = 2.5 $$ Since the limit of the expression inside the greatest integer function is $2.5$, and the greatest integer function is discontinuous at non-integer values, we must consider the behavior of the function as $x \to 0$. The function $\frac{3}{2} + \frac{x^2}{\sin x \tan x}$ approaches $2.5$ from values greater than $2.5$ or less than $2.5$. As $x \to 0$, $\frac{x^2}{\sin x \tan x} = \left(\frac{x}{\sin x}\right)^2 \cos x$. For $x$ near $0$ (but not $0$), $\frac{\sin x}{x} < 1$, so $\frac{x}{\sin x} > 1$. Thus $\left(\frac{x}{\sin x}\right)^2 > 1$. Also, for $x$ near $0$ (but not $0$), $\cos x < 1$. The product $\left(\frac{x}{\sin x}\right)^2 \cos x$ is slightly greater than $1$ for $x \neq 0$. Let $f(x) = \frac{x^2}{\sin x \tan x}$. We know $\lim_{x \to 0} f(x) = 1$. For $x \in (-\pi/2, \pi/2)$ and $x \neq 0$, we have $\sin x < x$ and $\tan x > x$. So $\sin x \tan x < x^2$. This implies $\frac{x^2}{\sin x \tan x} > 1$ for $x \in (-\pi/2, \pi/2)$ and $x \neq 0$. Therefore, as $x \to 0$, the term $\frac{x^2}{\sin x \tan x}$ approaches $1$ from values greater than $1$. So, $\frac{3}{2} + \frac{x^2}{\sin x \tan x}$ approaches $1.5 + 1 = 2.5$ from values greater than $2.5$. Let $y(x) = \frac{3}{2} + \frac{x^2}{\sin x \tan x}$. As $x \to 0$, $y(x) \to 2.5^+$. Thus, $\lim_{x \to 0} [y(x)] = [2.5^+] = 2$. The final value is $2$.